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Q.Using Kirchhoff's rules, calculate the potential difference between B and D in the circuit diagram as shown in the figure.

Tripura TbseCBSE Class XII Board 2018Subjective· 3mImportance★★★★★
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Bridge circuit for the Kirchhoff's-rules problem: three parallel branches between B and D (2 ohm+2 V then 1 ohm+1 V via A; 3 ohm+3 V then 1 ohm+1 V via C; a 2 ohm diagonal with no cell).
Bridge circuit for the Kirchhoff's-rules problem: three parallel branches between B and D (2 ohm+2 V then 1 ohm+1 V via A; 3 ohm+3 V then 1 ohm+1 V via C; a 2 ohm diagonal with no cell).

Reduce to three parallel branches between B and D; using VB−VD=∑εk/Rk∑1/RkV_B-V_D=\dfrac{\sum \varepsilon_k/R_k}{\sum 1/R_k} the numerator is zero, so VB−VD=0V_B-V_D=0.

Concept. Corner A connects only to B (via AB) and D (via AD), and corner C connects only to B (via BC) and D (via DC). So A and C are simple pass-through nodes, and the circuit is three branches in parallel between nodes B and D:

  • Branch via A: R=2+1=3 ΩR=2+1=3\,\Omega, cells 2 2\,V and 1 1\,V.
  • Branch via C: R=3+1=4 ΩR=3+1=4\,\Omega, cells 3 3\,V and 1 1\,V.
  • Diagonal: R=2 ΩR=2\,\Omega, no cell.

Kirchhoff's rules. Let V=VB−VDV=V_B-V_D. With each cell taken to drive current around the loop A→\toB→\toC→\toD in the same sense, the net emf aiding the D→\toB direction is +3 +3\,V in the A-branch and −4 -4\,V in the C-branch. Each branch current (taken D→\toB) is ik=εk−VRki_k=\dfrac{\varepsilon_k-V}{R_k}. KCL at node D (no external connection) gives ∑ik=0\sum i_k=0: …

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